**Proof.**
The fibres of $c$ are geometrically connected by More on Morphisms, Theorem 37.53.4. In particular $c$ is surjective. There are finitely many closed points $y = y_1, \ldots , y_ r$ of $Y$ where $X_ y$ has dimension $1$ and over $Y \setminus \{ y_1, \ldots , y_ r\} $ the morphism $c$ is an isomorphism. Some details omitted; hint: outside of $\{ y_1, \ldots , y_ r\} $ the morphism $c$ is finite, see Cohomology of Schemes, Lemma 30.21.1.

Let us carefully construct a map $b : c^*\omega _ Y \to \omega _ X$. Denote $f : X \to \mathop{\mathrm{Spec}}(k)$ and $g : Y \to \mathop{\mathrm{Spec}}(k)$ the structure morphisms. We have $f^!k = \omega _ X[1]$ and $g^!k = \omega _ Y[1]$, see Lemma 53.4.1 and its proof. Then $f^! = c^! \circ g^!$ and hence $c^!\omega _ Y = \omega _ X$. Thus there is a functorial isomorphism

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{F}, \omega _ X) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ Y)}(Rc_*\mathcal{F}, \omega _ Y) \]

for coherent $\mathcal{O}_ X$-modules $\mathcal{F}$ by definition of $c^!$^{1}. This isomorphism is induced by a trace map $t : Rc_*\omega _ X \to \omega _ Y$ (the counit of the adjunction). By the projection formula (Cohomology, Lemma 20.54.2) the canonical map $a : \omega _ Y \to Rc_*c^*\omega _ Y$ is an isomorphism. Combining the above we see there is a canonical map $b : c^*\omega _ Y \to \omega _ X$ such that

\[ t \circ Rc_*(b) = a^{-1} \]

In particular, if we restrict $b$ to $c^{-1}(Y \setminus \{ y_1, \ldots , y_ r\} )$ then it is an isomorphism (because it is a map between invertible modules whose composition with another gives the isomorphism $a^{-1}$).

Choose $m \in \mathbf{Z}$ as in (3) consider the map

\[ b^{\otimes m} : \Gamma (Y, \omega _ Y^{\otimes m}) \longrightarrow \Gamma (X, \omega _ X^{\otimes m}) \]

This map is injective because $Y$ is reduced and by the last property of $b$ mentioned in its construction. By Riemann-Roch (Lemma 53.5.2) we have $\chi (X, \omega _ X^{\otimes m}) =\chi (Y, \omega _ Y^{\otimes m})$. Thus

\[ \dim _ k \Gamma (Y, \omega _ Y^{\otimes m}) \geq \dim _ k \Gamma (X, \omega _ X^{\otimes m}) = \chi (X, \omega _ X^{\otimes m}) \]

and we conclude $b^{\otimes m}$ induces an isomorphism on global sections. So $b^{\otimes m} : c^*\omega _ Y^{\otimes m} \to \omega _ X^{\otimes m}$ is surjective as generators of $\omega _ X^{\otimes m}$ are in the image. Hence $b^{\otimes m}$ is an isomorphism. Thus $b$ is an isomorphism.
$\square$

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